2015 AMC 10A Problem 18

First locate $1000$ in hexadecimal: $1000 = 3 \cdot 256 + 14 \cdot 16 + 8$, so $1000 = 3E8_{16}$. Every integer from $1$ to $1000$ has at most three hex digits, with leading digit at most $3$.

Now count hex strings $d_2 d_1 d_0$ (leading zeros allowed) using only digits $0$ through $9$ whose value is at most $1000$. The leading digit $d_2$ must be $0, 1, 2$ or $3$, since $4$ would give at least $4 \cdot 256 = 1024 > 1000$. If $d_2 \le 3$ and the other digits are at most $9$, the largest such number is $399_{16} = 3 \cdot 256 + 9 \cdot 16 + 9 = 921 < 1000$, so every one of these strings is in range.

That gives $4 \cdot 10 \cdot 10 = 400$ strings, one of which is $000$, representing $0$, which is not a positive integer. Hence $n = 399$, and its digit sum is $3 + 9 + 9 = 21$.

The answer is $\boxed{\textbf{(E)}\ 21}$.

Counting numbers with restricted digits is a multiplication-principle count, but the upper limit must be handled carefully: convert the limit to the base in question, then check whether the largest "allowed" string falls below or above it. Here the check is clean ($399_{16} < 3E8_{16}$), so no boundary casework is needed. Also note the problem asks for the digit sum of $n$, not $n$ itself.

• Counting $400$ by including $0$ (which is not a positive integer), or not checking that $399$ in hex is really below $1000$ and $400$ in hex is above it.
• Answering $n = 399$ or its decimal-digit sum of the wrong quantity (for example summing the digits of $400$ to get $4$).